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Alexander And Hund’s Rule
By
Ian Beardsley
Copyright © 2021 Ian Beardsley
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Overview
What is meant by AI and Biological life chemistry are mathematical constructs?
We guess that artificial intelligence (AI) has the golden ratio, or its conjugate in its means geometric,
harmonic, and arithmetic by molar mass by taking these means between doping agents phosphorus (P)
and boron (B) divided by semiconductor material silicon (Si) :
The Golden Ratio
and.
Which can be written
We see that the biological elements, H, N, C, O compared to the AI elements P, B, Si is the golden ratio
conjugate (phi) as well:
We could begin with semiconductor germanium (Ge) and doping agents gallium (Ga) and Phosphorus (P)
and we get a similar equation:
,
In grams per mole. Then we compare these molar masses to the molar masses of the semiconductor
material Ge:
a
b
(
Φ
)
a
b
=
b
c
a = b + c
Φ = 1/ϕ
2PB
P + B
1
Si
=
2(30.97)(10.81)
30.97 + 10.81
1
28.09
= 0.57
0.65 + 0.57
2
= 0.61 ≈ ϕ
PB(P + B) + 2PB
2(P + B)Si
≈ ϕ
C + N + O + H
P + B + Si
≈ ϕ
2G a P
G a + P
= 42.866
G a P = 46.46749
2G a P
G a + P
1
G e
=
42.866
72.61
= 0.59
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Then, take the arithmetic mean between these:
We then notice this is about the golden ratio conjugate, , which is the inverse of the golden ratio, .
. Thus, we have
1.
2.
Another relationship to biological life where AI is concerned. CHNOPS are another categorization of
biological elements in biology:
One has to ask: Do first generation semiconductors which were germanium, and second generation
semiconductors (which is where we are today) have a function if we use them together. After all..,.
We can write
This is the quadratic
Which has solutions
G a P
1
G e
=
46.46749
72.61
= 0.64
0.59 + 0.64
2
= 0.615
ϕ
Φ
ϕ ≈
1
Φ
G a P(G a + P) + 2GaP
2(G a + P )G e
≈ ϕ
G a P(G a + P) + 2GaP
2(G a + P )Si
≈ Φ
G a As(G a + A s) + 2G a As
2(G a + A s)G e
≈ 1
C + H + N + O + P + S
G a + A s + G e
≈
1
2
G e − Si
Si
= Φ
1
Ge
Si
= ϕ
x
2
− 3x y + y
2
= 0
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We can say…
With an accuracy of 72.64/73.54=98.8%
Apparently electrical engineers are working on silicon-germanium hybrids for a possible next generation
semiconductor.
y =
(
3
2
−
5
2
)
x = 0.381966x
y =
(
3
2
+
5
2
)
x = 2.618033989x
G e = (Φ + 1)Si
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Alexander And Hund’s Rule
I wrote a paper about AI and Biological elements being mathematical constructs that are
related to one another. Александр Сергеевич of Institute of Physical Chemistry (Moscow)
wrote /
Your approach is very relevant and touches on one of the main provisions of quantum
chemistry-namely, the well-known Hund’s rule that sets the number of electrons involved in the
formation of a bond (usually n=2). However, it is still not clear why only two electrons must
necessarily participate in the formation of a chemical bond - there are no prohibitions in
quantum mechanics in this case! Perhaps your proposed classification confirms the possibility
of participation in the formation of a chemical bond of several electrons, more than two in
number! Then it is possible to substantiate the presence of some criterion that makes it
possible to distinguish between biologically active and passive elements. And this is extremely
important for understanding how life arose on Earth!
I have begun to approach this by looking at the simplest possible molecule, which would be
hydrogen gas, H2 by looking at it as a mathematical and geometric concept. I find this resolves
as a sphere minus a hyperboloid with Gaussian distributions as inlets. Such an example
indicates that Hund’s rule may be connected to a need for symmetry which can be achieved by
bonds with two electrons. In my earlier mathematical constructs I used molar masses, whose
pertinence I see as affecting the geometry of the substance because it determines the motions
available to the electrons in the lattice./
Ruud Loeffen wrote to Александр Сергеевич:/
I am very glad Ian found somebody who understands his work. It seems very important to me. I
hope you two will go on cooperating! Good luck!
To which Александр Сергеевич responded:
This is an interesting proposal. But I was fond of quantum chemistry in my “green” years. And
these ideas had visited me a that time, more than 40 years ago. I am afraid that I am too old for
this adventure! But I do not refuse possible cooperation. If not me, then my students will
continue my work! Let’s go to work, gentlemen!
Let’s go to work! We begin with the equation of a hyperbola:/
/
Its volume is as a solid of revolution:/
/
The Equation of a circle is:/
/
(
y
a
)
2
−
(
x
b
)
2
= 1
V = π
a
2
b
2
∫
b
a
(
b
2
+ x
2
)
d x = π
a
2
b
2
(
b
2
x +
x
3
3
)
{
x 2
x1
(
x − a
)
2
+
(
y − b
)
2
= r
2
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And its volume as a solid of revolution is:/
/
The equation of a gaussian distribution is:/
/
f(x)=probability distribution, sigma=standard deviation, mu=mean/
Now see why I call upon the hyperboloid, sphere, and gaussian distribution, examine the
following sketches…1
V =
4
3
πr
3
f (x) =
1
σ 2π
e
−
1
2
(
x − μ
σ
)
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To do our calculations we will leave out the inlets as they are nominal. The radius of diatomic
hydrogen is 120 picometers (120 pm). The volume of the hyperboloid is/
/
x1=-30 pm and x2=30 pm and the bohr radius, which is the most likely, or frequent electron-
proton separation is 52.9177 pm. This is the value of a. The volume of the sphere is:/
/
Subtracting the volume of the hyperboloid from the volume of the sphere we have (if a=b):/
/
V
H
= π
(
a
b
)
2
(
b
2
x
2
+
x
2
3
3
)
− π
(
a
b
)
2
(
b
2
x
1
+ x
1
3
/3
)
= 188.5a
2
pm + 9000
a
2
b
2
2π
4
3
πr
3
=
4
3
60
3
= 904,778.684pm
3
V = V
S
− V
H
= 904,778.684 − 584402.6842 = 320376pm
3
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Taking the cube root we have/
/
This is very close to the midpoint between the radius of hydrogen (53pm) and H-H bond length
which is 74 picometers. Which is 63.5pm. It becomes clear what the drawing is:/
1
3
320376 = 68.4pm
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Considering methane (CH4) with a diameter of 400pm we have:/
/
With a=b=bohr radius./
The area of the sphere is:/
/
/
/
The bond length of CH4 is 109pm/
/
/
See illustrations on next page…1
V
H
= 2π
a
2
b
2
(
b
2
x +
x
2
3
)
{
200
0
= 400πa
2
+
a
2
b
2
+
a
2
b
2
(8E6)/3 = 20274100pm
3
V
s
=
4
3
πr
3
=
4
3
π 400
3
= 268082573.1pm
3
V
s
− V
H
= 266407012pm
3
3
x = 643.45pm
2(400) + 218 = 1018
ϕ(1018) = 630pm ≈ 643.45pm
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The Author/